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Amiraneli [1.4K]
3 years ago
10

Round off the measurement to three significant figures 12.17º C

Chemistry
1 answer:
prohojiy [21]3 years ago
6 0
12.2 C
It has 3 significant figures now.
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I have no idea how to do this and need help
den301095 [7]

Answer:

60.55%

Explanation:

nO3=155/48

nO2 used: 3/2nO3=4.84375

percent yield: 4.84375/8=60.55%

5 0
3 years ago
If the gold in the crown was mixed with a less-valuable metal like bronze or copper, how would that affect its density?
Sindrei [870]

Answer: If the gold in the crown was mixed with a less-valuable metal such as bronze of copper then this affects its density by making it to weigh even more and if it weighs more then it has less of chance to float. Observe: Drag each of the crowns into the liquid.

Explanation:

6 0
3 years ago
A compound is found to contain 38.65 % carbon, 16.25 % hydrogen, and 45.09 % nitrogen by mass. what is the empirical formula for
iren [92.7K]

The molar mass of carbon is 12, hydrogen is 1, and nitrogen is 14, hence the ratio are:

 

C = 38.65 / 12 = 3.22

H = 16.25 / 1 = 16.25

N = 45.09 / 14 = 3.22

 

Divide the three by the lowest ratio which is 3.22:

 

C = 3.22 / 3.22 = 1

H = 16.25 / 3.22 = 5

N = 3.22 / 3.22 = 1

 

So the empirical formula is:

CHN

5 0
3 years ago
Nitrogen dioxide and water react to form nitric acid and nitrogen monoxide, like this:
posledela

Answer: The value of the equilibrium constant Kc for this reaction is 3.72

Explanation:

Equilibrium concentration of HNO_3 = \frac{15.5g}{63g/mol\times 9.5L}=0.026M

Equilibrium concentration of NO = \frac{16.6g}{30g/mol\times 9.5L}=0.058M

Equilibrium concentration of NO_2 = \frac{22.5g}{46g/mol\times 9.5L}=0.051M

Equilibrium concentration of H_2O = \frac{189.0g}{18g/mol\times 9.5L}=1.10M

Equilibrium constant is defined as the ratio of concentration of products to the concentration of reactants each raised to the power their stoichiometric ratios. It is expressed as K_c  

For the given chemical reaction:

2HNO_3(aq)+NO(g)\rightarrow 3NO_2(g)+H_2O(l)

The expression for K_c is written as:

K_c=\frac{[NO_2]^3\times [H_2O]^1}{[HNO_3]^2\times [NO]^1}

K_c=\frac{(0.051)^3\times (1.10)^1}{(0.026)^2\times (0.058)^1}

K_c=3.72

Thus  the value of the equilibrium constant Kc for this reaction is 3.72

5 0
2 years ago
¿Que atomo gana el electron?
antoniya [11.8K]

Answer:

anion

Explanation:

3 0
3 years ago
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